Results 1 to 8 of 8

Math Help - differentiation

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    40

    differentiation

    I need help with the following derivatives

    y= (e^1/x) / x^2


    f(t) = t ln(5t)

    thanks for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    Could you show us what exactly is giving you trouble? What do you get when you attempt them?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by tim_mannire View Post
    I need help with the following derivatives

    y= (e^1/x) / x^2


    f(t) = t ln(5t)

    thanks for your help
    USE QUOTIENT RULE ON THE FIRST ONE AND PRODUCT rULE ON THE SECOND ONE. SHOW YOUR WORK IF YOU GET STUCK!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    40
    Using the rules you suggested i got

    dy/dx = ((x^2)(e^1/x) - (e^1/x)(2x)) / x^4

    f ' (t) = (1/5t) (t)


    are they correct?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by tim_mannire View Post
    Using the rules you suggested i got

    dy/dx = ((x^2)(e^1/x) - (e^1/x)(2x)) / x^4

    f ' (t) = (1/5t) (t)


    are they correct?
    For dy/dx it appears you've not differentiated e^{1/x} correctly

    It's easier if you sub u = \frac{1}{x} and then use the chain rule.

    --------------------------------------------------

    For the second one remember the chain rule

    \frac{d}{dx} \ln [f(x)] = \frac{f'(x)}{f(x)}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jun 2008
    Posts
    40
    thanks for your help. I have tried what you suggested but I just don't understand . A complete working of the question using the rules would help me to understand.

    thank you
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by tim_mannire View Post
    thanks for your help. I have tried what you suggested but I just don't understand . A complete working of the question using the rules would help me to understand.

    thank you
    y = \frac{e^{1/x}}{x^2}

    For this we need to use the quotient rule which is: y' = \frac{vu' - uv'}{v^2}

    In your case u = e^{1/x} and v = x^2

    I will differentiate the numerator as that's what seems to be the issue then you can try using my value of u' in the quotient rule. FYI my final answer is \frac{dy}{dx} = \frac{e^{1/x}(2x-1)}{x^4} = -\frac{e^{1/x}(1-2x)}{x^4}


    To find u' (du/dx) we need to work out \frac{d}{dx} e^{1/x}.
    For simplicity and clarity's sake let t = \frac{1}{x}. From the chain rule we know that \frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx}

    Due to this substitution we can rewrite u as u = e^t.

    Differentiating this function is easy: \frac{du}{dt} = e^t

    Similarly 1/x is also easy to differentiate: \frac{dt}{dx} = -\frac{1}{x^2}

    From the chain rule above: u' = \frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx} = e^t \cdot \frac{1}{x^2}

    Since t = \frac{1}{x} we get u' in terms of x:

    Therefore u' = -\frac{1}{x^2}\,e^{1/x}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    Quote Originally Posted by tim_mannire View Post
    f(t) = t ln(5t)
    For this one, it is

    f'(t) = t \cdot [\ln(5t)]' + [t]' \cdot \ln(5t)

    = t \cdot \frac{5}{5t} + 1 \cdot \ln(5t)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. Differentiation and partial differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 30th 2010, 10:16 PM
  3. Differentiation
    Posted in the Calculus Forum
    Replies: 10
    Last Post: February 9th 2010, 01:03 AM
  4. Differentiation and Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 6th 2009, 04:07 AM
  5. Differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 5th 2008, 06:09 AM

Search Tags


/mathhelpforum @mathhelpforum